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The partition function is
$$Z_N = \sum_{\sigma_1 = \pm 1} \sum_{\sigma_2 = \pm 1} ... \sum_{\sigma_N = \pm 1} \exp\{ K \sigma_1 \sigma_2 + H \sigma_1 \} \exp\{ K \sigma_2 \sigma_3 + H \sigma_2 \} ... \exp\{ K \sigma_N \sigma_1 + H \sigma_N \}$$ 

We postulate the existence of an operator $T$ such that the above can be rewritten
$$Z_N = \sum_{\sigma_1 = \pm 1} \sum_{\sigma_2 = \pm 1} ... \sum_{\sigma_N = \pm 1} \langle \sigma_1 | T | \sigma_2 \rangle \langle \sigma_2 | T | \sigma_3 \rangle ... \langle \sigma_N | T | \sigma_1 \rangle$$
where the matrix elements should evaluate to e.g.
$$\langle +1 | T | -1 \rangle = \exp\{ K (+1)(-1) + H (+1) \} = e^{-K+H}$$
and similarly for all 4 spin combinations of $\sigma_i = \pm 1$, $\sigma_{i+1} = \pm 1$.

Now since in general
$$\sum_n | n \rangle \langle n | = I$$
i.e. the sum of projection operators is the identity operator, we can sum out all but the first and last bra and ket from the partition function, to give
$$ Z_N = \sum_{\sigma_1 = \pm 1} \langle \sigma_1 | T^N | \sigma_1 \rangle $$
where $T^N$ stands for repeated application of $T$.

By choosing a vector representation for $| \pm 1 \rangle$ we can find a matrix representation of $T$. We make the arbitrary choice 

$$\langle +1 | = (1,0), \langle -1 | = (0,1) \Rightarrow | +1 \rangle = (1,0)^\dagger, | -1 \rangle = (0,1)^\dagger $$

and simply by testing possible combinations of vectors, we find the following matrix representation

\begin{eqnarray*}
T = \left(
\begin{array}{cc}
  e^{K+H} & e^{-K+H} \\
  e^{-K-H} & e^{K-H}
\end{array}
\right)
\end{eqnarray*}

Hence the partition function has the interpretation of a trace over $T$

$$Z_N = \mathrm{Tr}( T^N )$$

To evaluate this trace, assume that $T$ can be diagonalised, i.e. there exists a matrix $\Lambda$ such that

$$\Lambda T \Lambda^{-1} = S = \mathrm{diag}( \lambda_i )$$ and $$\Lambda \Lambda^{-1} = I$$

where $S$ is a matrix with the eigenvalues of $T$ as its diagonal entries, and zeroes everywhere else. Then we have

$$T = \Lambda^{-1} S \Lambda \Rightarrow T^N = \Lambda^{-1} S \Lambda \Lambda^{-1} S \Lambda ... \Lambda^{-1} S \Lambda = \Lambda^{-1} S^N \Lambda$$

and since $\mathrm{Tr}(AB) = \mathrm{Tr}(BA)$ and $S^N = \mathrm{diag}(\lambda^N_i)$ we have

$$Z_N = \mathrm{Tr}(\Lambda^{-1} S^N \Lambda) = \mathrm{Tr}(S^N) = \sum_i \lambda_i^N$$

Now since $T$ is positive definite (proof?), $\lambda_i > 0$, i.e. all eigenvalues are positive, the asymptotic behaviour

$$Z_N \sim_{N \rightarrow \infty} \lambda_+^N$$

where $\lambda_+$ is the largest eigenvalue.

To find the eigenvalues of $T$ we find the roots of the characteristic polynomial

$$ (e^{K+H} - \lambda) (e^{K-H} - \lambda) - e^{-2K} = \lambda^2 - 2 e^K \cosh H \lambda + 2 \sinh 2K = 0 $$

$$\Rightarrow \lambda = e^K \cosh H \pm \sqrt{ e^{2K} \cosh^2 H - 2 \sinh 2K }$$

Then if we define

$$x = \sqrt{ 1 - \frac{2 \sinh 2K}{e^{2K} \cosh^2H} }$$

the partition function becomes

$$Z_N = e^{NK} \cosh^N H [ (1 + x)^N + (1 - x)^N ]$$

and in the thermodynamic limit is given by the dominant eigenvalue, raised to $N$

$$Z_N = e^{NK} \cosh^N H [ 1 + x ]^N$$

The Helmholtz free energy is

$$\mathcal{F} = - k_B T \log Z_N = - N k_B T \left[ K + \log \cosh H + \log ( 1 + x) \right]$$

and in the limit $H \rightarrow 0$ we replace $\cosh 0 = 1$ and $x(H=0) = e^{-2K}$ to obtain

$$Z_N(H=0) = [ e^K + e^{-2K} ]^N = ( 2 \cosh K )^N$$

i.e. we recover the result with zero applied field in the Hamiltonian, with a free energy

$$\mathcal{F}(H=0) = - N k_B T \log( 2 \cosh K)$$

The average magnetisation

$$m = \frac{1}{N} \frac{1}{Z_N} \frac{ \partial }{ \partial H } Z_N = \frac{1}{N} \frac{1}{\lambda_+^N} \frac{ \partial }{ \partial H } ( \lambda_+^N ) = \frac{1}{\lambda_+} \frac{ \partial }{ \partial H } \lambda_+ $$

consider just the derivative

$$\frac{ \partial }{ \partial H } \lambda_+ = e^K \sinh H ( 1 + 1 / x )$$

hence

$$m = \frac{ e^K \sinh H ( 1 + 1 / x ) }{ e^K \cosh H ( 1 + x ) } = \tanh H / x$$

and in the absence of an applied field we recover $m(H=0) = 0$ for all positive $K$; as expected there is no ordered phase in 1 dimension with no applied field (no spontaneous symmetry breaking) at finite temperature.

To find the susceptibility at zero applied field we must first derive again, but we can simplify this procedure by noting that all positive powers of $H$ will vanish in the limit $H \rightarrow 0$, so it suffices to expand $m$ to first order in $H$ before deriving

$$m \approx \frac{H}{ \sqrt{ 1 - 2 e^{-2K} \sinh 2K } }  = He^{2K} \Rightarrow \left. \frac{ \partial m }{ \partial H } \right|_{H=0} = e^{2K}$$

Large fluctuations in the order parameter (divergence of the susceptibility of the parameter) are characteristic of a phase transition. For the 1D Ising magnet we see that the susceptibility diverges at $K \rightarrow \infty \Rightarrow T \rightarrow 0$, so there is no phase transition at finite temperature.

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